Animated Solution for Mathematics - Sequence and Series: If $a,b,c,d$ are positive real numbers such that $a+b+c+d=2$, then $M=(a+b)(c+d)$ satisfies the relation

Imagine you are standing before a problem that seems designed to confuse you. You have four variables: $a$, $b$, $c$, and $d$. They are all positive real numbers, and they are bound by the constraint $a+b+c+d=2$.

Your goal is to find the range of the expression $M=(a+b)(c+d)$. At first glance, this looks like a mess. How do you handle four variables simultaneously?

The secret, as in many JEE problems, is not to fight the complexity, but to simplify it. Let us perform a clever substitution. Let $x=a+b$ and $y=c+d$.

Suddenly, the expression $M=(a+b)(c+d)$ transforms into the much friendlier:

We have successfully reduced a four-variable problem into a two-variable problem. This is the first step in mastering the JEE Advanced approach: look for symmetry and structure.

Now, let us look at our constraint. We know that $a+b+c+d=2$. With our new variables, this becomes:

Since $a,b,c,d$ are all strictly positive, it follows that $x$ and $y$ must also be strictly positive. If you were to visualize this on a Cartesian plane, you are looking at a line segment in the first quadrant where $x+y=2$.

Every point $(x,y)$ on this line represents a possible configuration of our original variables. Our target expression $M=xy$ is the area of a rectangle with sides $x$ and $y$.

As the point $(x,y)$ moves along the line $x+y=2$, the rectangle changes shape, and its area $M$ changes accordingly. We are essentially asking: what is the range of the area of a rectangle whose top-right corner is constrained to a specific line?

This is where we bring out the heavy artillery. Whenever you have a fixed sum of positive terms and you want to analyze their product, the Arithmetic Mean-Geometric Mean (AM-GM) inequality is your best friend.

The theorem states that for any positive real numbers $x$ and $y$, the Arithmetic Mean is always greater than or equal to the Geometric Mean:

This inequality is elegant, powerful, and perfectly suited for this problem. We know the sum $x+y=2$. Let us substitute this into the inequality:

Simplifying the left side, we get $1\ge \sqrt{xy}$. Since $M=xy$, this becomes $1\ge \sqrt{M}$.

We are almost there. To isolate $M$, we square both sides of the inequality $1\ge \sqrt{M}$. Since $M$ is an area and must be positive, we can safely square both sides without worrying about the inequality sign:

This tells us the maximum value of $M$ is $1$. But what about the lower bound? We know $x>0$ and $y>0$, so their product $M=xy$ must be strictly greater than $0$.

Combining these two insights, we get $0<M\le 1$. This matches the interval $0<M\le 1$ provided in the options.

You have just navigated a complex algebraic problem using nothing but logical grouping and a fundamental inequality. This is the essence of JEE mathematics: finding the most elegant path to the truth.